# Thread: Metric space question

1. ## Metric space question

Can someone help with this please, i'm stuck;

X = [0,∞) and d(x,y) = │x-y│/1+│x-y│

Show that it's a Metric space; I've done the first two conditions but stuck proving the triangle inequality...

Then it says describe B1(0) the ball of centre 0 and radius 1...surely there's not much to say?

2. $\frac{{\left| {x - y} \right|}}{{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {x - y} \right|}}{{1 + \left| {x - y} \right|}} = d(x,y)$

$\frac{{\left| {y - z} \right|}}{{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {y - z} \right|}}{{1 + \left| {y - z} \right|}} = d(y,z)$

$d(x,z) = \frac{{\left| {x - z} \right|}}
{{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {x - y} \right| + \left| {y - z} \right|}}
{{1 + \left| {x - y} \right| + \left| {y - z} \right|}}$

$d(x,z) \leqslant \frac{{\left| {x - y} \right| + \left| {y - z} \right|}}
{{1 + \left| {x - y} \right| + \left| {y - z} \right|}} \leqslant \frac{{\left| {x - y} \right|}}
{{1 + \left| {x - y} \right|}} + \frac{{\left| {y - z} \right|}}
{{1 + \left| {y - z} \right|}} = d(x,y) + d(y,z)$